Let $\alpha$ be an ordinal , and define $X_{\alpha} = \alpha \times [0,1)$ to be a topological space with the lexicographic order.
Denote by $\omega$ the first countable ordinal and $\Omega$ the first ordinal that is not countable.
I need to prove the following :
if $\alpha$ is countable then $X_{\alpha}$ is isomorphic to $[0,1)$ as ordered set and to conclude that $X_{\alpha}$ is homeomorphic to $[0,1)$.
and, for each point in $X_{\Omega}$ except the min. element there is a neighborhood that homeomorphic to $R$.
Im a bit stuck on this question, for the first part i thought sending $(n,x)$ to $x$ but this does not work. i tried also sending $(n,x)$ to anywhere in the interval $(1/(n+1) , 1/n)$ - this also does not work.
I would really appreciate a detailed answer , as im not familiar that much with ordinals.
Thanks for helping.
We can use transfinite recursion to inject $\alpha$ into $[0,1)$ such that between any two elements of the image of $\alpha$ there are uncountably many real numbers. As pointed out in the comments, this is easily established even without transfinite induction. In fact, every countable linear order embeds into the rationals, so you can just use that embedding.
However you obtain such a function, call it $f$. We then create $g:\alpha\times [0,1)\to [0,1)$ by saying that $g(x,0)=f(x)$ and observing that since $[f(x),f(s(x)))$ is an open interval, there's a natural order preserving bijection between it an $[0,1)$ which we use to assign the value of $g(x,y)$ for $y\neq 0$. This natural bijection sends $[a,b)$ to $[0,1)$ via the formula $h(x)=(x-a)/b$